Question

In part a, you found the amount of product (3.80 mol p2o5 ) formed from the given amount of phosphorus and excess oxygen. in part b, you found the amount of product (3.40 mol p2o5 ) formed from the given amount of oxygen and excess phosphorus. now, determine how many moles of p2o5 are produced from the given amounts of phosphorus and oxygen. express your answer to three significant figures and include the appropriate units. view available hint(s)

Answer 1

The balanced chemical equation between phosphorus and oxygen can be given as,

`4 P(s) + 5 O_(2)(g) -->2 P_(2)O_(5)(s)`
`8.50 mol O_(2) * (2 mol P_(2)O_(5))/(5 mol O_(2)) = 3.40 mol P_(2)O_(5)`

3.80 mol
`P_(2)O_(5)` is formed from P (limiting reactant):

Moles of P =
`3.80 mol P_(2)O_(5) * (4 mol P\pi)/(2 mol P_(2)O_(5)) = 7.6 mol P`

3.40 mol
`P_(2)O_(5)` is produced when Oxygen is limiting:

Moles of
`O_(2)` =
`3.40 mol P_(2)O_(5) * (5 mol O_(2))/(2 mol P_(2)O_(5)) = 8.5 mol O_(2)`

So we have 7.6 mol P and 8.5 mol O2.

Calculating the moles of P2O5 formed from the given moles of P and O2:

Moles of phosphorus pentoxide from P =
`7.60 mol P * (2 mol P_(2)O_(5))/(4 mol P)`

So the moles of
`P_(2)O_(5)` produced from P are

=
`3.80 mol P_(2)O_(5)`

Moles of phosphorus pentoxide formed from O2 =
`8.5 mol O_(2) * (2 mol P_(2)O_(5))/(5 mol O_(2)) = 3.40 mol P_(2)O_(5)`

O2 produced the least amount of the product so oxygen would be the limiting reactant.

Moles of
`P_(2)O_(5)` produced will be 3.40 mol